Dimension Computations in Non-Commutative Associative Algebras

Transcription

1 Dimension Computations in Non-Commutative Associative Algebras by Grischa Studzinski Diploma thesis in mathematics submitted to Fakultät für Mathematik, Informatik und Naturwissenschaften der Rheinisch-Westfälischen Technischen Hochschule Aachen May 9, 2010 produced at Lehrstuhl D für Mathematik Prof. Dr. Eva Zerz

2 Introduction Any finitely generated associative algebra can be presented as a factor of the free associative algebra. Therefore computations in the free algebra have many applications in different areas of mathematics, like cryptography, ring theory, homological algebra, representation theory of monoids, groups and algebras, algebraic system and control theory, quantum algebras, in mathematical and theoretical physics. The aim of this diploma thesis is to study factors of the free algebra with focus on the K-dimension. In particular, we want to answer the question whether a factor algebra, given by a two-sided ideal, is finite dimensional or not. Here the approach to answer this question is to gather information which hopefully solves the question by studying the Gröbner basis. Therefore one needs to construct a Gröbner basis explicitly from a given set of generators for an ideal. In theory this question was studied since the early years of computer algebra: Mora ([Mor86, Mor89, Mor94]), E. Green ([Gre93, Gre00]), Ufnarovskij ([Ufn90, Ufn98]) and Cojocaru et al. ([CPU99]) presented different facets of what we call today non-commutative Gröbner basis theory. In particular Mora discussed free non-commutative algebras and their quotient rings endowed also with negative (non-well-)orderings and further extended his theory (with Apel, [Ape00]). Other important contributions were made by Apel and Lassner ([AL88]) and especially Apel in [Ape00]. In the last years there has been more progress in theoretical, implementational and practical directions. Notably, the interest in free associative algebras grew stronger, as indicated by e. g. the book of D. Green ([Gre03]), where the author considers also negative (non-well-)orderings for certain non-commutative cases with a very different motivation and meaning, compared to the theory of Mora ([Mor89]) and Apel ([Ape00]) and with the commutative case as in Greuel et al. ([GP02]). Evans and Wensley investigated in [EW07] involutive bases in non-commutative algebras. With the recent work of La Scala and Levandovskyy [LL09] a new way to compute Gröbner bases was born, where non-commutative Gröbner bases of graded ideals in free algebras are computed via the Letterplace correspondence. The most important point for practical computer algebra is that the computations take place in a commutative ring, where the data structures as well as many fundamental algorithms have been deeply studied and enhanced in the past 40 years. 2

3 So another task is to translate the setup of computing the K-dimension for factor algebras into the realm of Letterplace. Here we found some interesting aspects, as well as new structures, as for example the K-shift-basis (see 2.59). Along the way there were many interesting applications and some theoretical development, like the mistletoes (see 2.43), which are a completely new way to store bases for factor algebras compactly, the concept of fake dimension (see 2.13) and the usage of the Ufnarovskij graph to determine the finiteness of the factor algebra by a given truncated Gröbner basis (see 2.19), which found their way into this thesis. The mistletoes resemble the concept of border bases (cf. [KK06]). However, the connection to border bases is still to be investigated deeper. The algorithms to compute mistletoes and the K-dimension have been analyzed for their algorithmic complexity(2.55,2.58). The usage of the Ufnarovskij graph allows one to detect early the finiteness of a K-basis, if applied in an adaptive algorithm for the computation of a Gröbner basis, what implies the finiteness of a Gröbner basis in this situation. Adaptive computation of a Gröbner basis -either in the classical or in the Letterplace setting- is examined and realistic bounds for a single adaptive step are established (1.53,1.54,). One of my personal goals for this thesis is to give an easy-to-understand introduction to non-commutative calculus in the free algebra, especially to noncommutative Gröbner bases, since although well studied, most work dealing with such general structures as the free associative algebra has not an introductive character. Moreover, this work is a starting point for applications of non-commutative methods relying on Gröbner bases in free associative algebras and we are planing to expand our methods to other fields, like Gröbner basis cryptosystems and computations in non-commutative modules (see for example [AK05] and [BK07]). Alongside the theoretical development we implemented the procedures in the computer algebra system Singular. Singular [GPS09] has been developed since more than 20 years under the direction of Prof. Greuel, Prof. Pfister and Dr. Schönemann in Kaiserslautern, Germany. Singular is a specialized computer algebra system for supporting research in commutative algebra, algebraic geometry and singularity theory. Since 2005, there is a subsystem Singular:Plural [GLS06], which provides Gröbner bases-related functionality for a class of non-commutative GR-algebras [Lev05]. Special data structures, developed and implemented for polynomials, together with carefully designed and implemented algorithms, contribute to the widespread acceptance of Singular as one of the fastest computer algebra systems in the world. The recently developed Letterplace paradigm allows the computation of Gröbner 3

4 bases in the free associative algebra and the corresponding algorithms have been implemented in the computer algebra system Singular. I would like to thank many people for their support. First of all there is my supervisor Dr. Viktor Levandovskyy, who gave me the opportunity for this work and a lot of help, who encouraged me on my way and of course for lots of interesting discussions during this work. Also I would like to thank my dear colleague Daniel Andres, who always was a great help and the one I could turn to with any questions, without annoying our supervisors. Of course my sincere thanks go to my correctors, Prof. Dr. Eva Zerz and Prof. Dr. Martin Kreuzer, for the time they spent to review my thesis and for their useful suggestions. Moreover, I would like to thank Prof. Dr. Roberto La Scala, Prof. Dr. Alexander Tiskin, Prof. Dr. Gunther Malle and Dr. Jürgen Müller, for their contributions by sharing their knowledge with me. Aachen, May 9, 2010 Grischa Studzinski 4

5 Contents 1 Non-Commutative Gröbner Bases Notations and Orders Gröbner Bases and Normal Forms The Gröbner Basis Algorithm The Diamond Lemma Truncated Gröbner Bases The Letterplace Approach Factor Algebras A Basis for a Factor Algebra Graphs and Trees The Dimension of Factor Algebras K-Bases of Factor Algebras Mistletoes Factor Algebras over Letterplace Rings Implementation The Data Structure Main Procedures Determine Finiteness of K-Dimension Harvesting the Mistletoes Determine K-Dimension Computing the Coefficients of the Hilbert series Combined Procedures Procedures Dealing with Mistletoes Determine K-Dimension Computing the Coefficients of the Hilbert series Other Procedures An Example in Singular Other Computer Algebra Systems Examples Explanation of the Examples Timings A Comparison to GAP Erklärung 71 5

6 1 Non-Commutative Gröbner Bases The goal of this section is to introduce Gröbner bases of ideals of the free algebra K X. Most of this chapter is basic knowledge and well studied (see for example [Ufn98], [Coh07], [GP02]). However, this knowledge is needed for a proper understanding of most computations in non-commutative algebras, and of great relevance for factor algebras. 1.1 Notations and Orders Throughout this chapter let K be a field and X be the free monoid on n generators, denoted by x 1,...,x n. We define the free algebra as the monoid ring K X := { i I α i m i α i K, m i X, I an arbitrary index set, only finitely many α i 0} and call the elements of K X polynomials and the elements of X embedded in K X together with the identity 1 monomials. A subset I K X is called (two-sided) ideal of K X, written I K X, if 1. 0 I 2. r, s I r + s I 3. r K X, s I rs I, sr I An ideal I is called proper, if I K X and I 0. A set G I is called generating set, written I = G, if s I g i G, r i, l i K X : s = l i,j g i r i,j. If there exists a finite generating set, I is called i j finitely generated. Since K is a field there is no loss of generality to assume that all polynomials of a given generating set are monic. 1.1 Definition. An (strict total) ordering < is a total, transitive and asymmetric relation on X, that is If a < b then (a > b) (asymmetry); If a < b and b < c then a < c (transitivity); Either a < b or b < a a, b X, a b (totality). 6

7 1.2 Definition. A total ordering < on X is called a well ordering, if every non-empty subset of X has a least element with respect to <. In particular, 1 < x x X. reduction ordering, if for all m 1, m 2, l, r X with m 1 < m 2 we have lm 1 r < lm 2 r. monomial ordering, if it is a well ordering and a reduction ordering. Note that for a reduction ordering we have if m, n X are such that n divides m, that is, if there exists l, r X with m = lnr, denoted by n m, then we have n < m, because for 1 < l, r X we have n = 1n < ln = ln1 < lnr = m. With a given ordering we can write each polynomial f K X uniquely as f = k i=1 c im i, such that c i K and m i X with m 1 < < m k. In this work we will always assume that < is a monomial ordering (for existence see 1.4). 1.3 Definition. Given an ordering <, we define the leading monomial of a polynomial f = k i=1 c im i 0 as the maximum (with respect to <) of the set {m i c i 0}. and denote it by lm(f). Also we call the coefficient of lm(f) the leading coefficient, denoted by lc(f) and we define the leading term of f as lt(f) = lc(f) lm(f). Finally we will denote with L( I ) the leading ideal of an ideal I, which is the ideal generated by the leading monomials of I. 1.4 Example. Without loss of generality, we can always assume that x 1 < x 2 <... < x n. Then we have the following two examples of monomial orderings: Let µ, ν X, such that µ = x j1 x j2 x jk, ν = x l1 x l2 x l k. Then we have: µ < lex ν 1 i min{k, k} : x jw = x lw w < i x ji < x li This is called the (left) lexicographical ordering. Take µ, ν as before. We define: µ < gradlex ν k < k or k = k and µ < lex ν. This is called the graded or degree lexicographical ordering. 1.5 Definition. For a given ordering < we define the multidegree of a monomial j m = x k 1 i 1 x k j i j as the k-tuple (k 1,...,k j ) and the total degree as k r. The (total) degree of a polynomial f is defined as the (total) degree of its leading monomial. We denote the total degree of f by deg t (f) and the multidegree by deg(f). r=1 7

8 1.2 Gröbner Bases and Normal Forms 1.6 Definition. Let G K X \{0} and G =: I. A normal form of f K X with respect to G is an element g K X such that f g I and either g = 0 or lm(g i ) lm(g) g i G. We denote a normal form of f with respect to G by NF(f,G). A subset G I is called a Gröbner basis of I if the leading monomial of an arbitrary element in I is a multiple of the leading monomial of an element in G. Equivalently, G is a Gröbner basis if {lm(g) g G} = L(I). 1.7 Remark. Note that a Gröbner basis always exists, since we can take G = I \ {0}. This is due to the fact that we do not demand our Gröbner basis to be finite. In fact there are some ideals, which do not posses a finite Gröbner basis, cf One can easily see the relevance of Gröbner bases: If G is a Gröbner basis of I then a normal form for f I is given by 0 and this is the only choice we have. However, neither the normal form nor the Gröbner basis are unique in general. 1.8 Example. If G is a Gröbner basis of an ideal I and G I, then G := G {g}, g I\G is again a Gröbner basis. Take B := {x 2 } x 2 K[x 1, x 2 ] with the degree lexicographical ordering with respect to x 1 > x 2 and consider f = x 1. Then g 1 = x 1 and g 2 = x 1 x 2 are both normal forms of f with respect to B. Note that B is already a Gröbner basis for x 2. Note that g 2 has terms which are contained in L(I), which is the reason why we have two different normal forms. 1.9 Definition. A normal form g = k a i t i, a i K, t i X of f K X with i=0 respect to G is called reduced, if g is monic, that is, its leading coefficient is 1, and if lm(g w ) t i i = 0,..., k, g w G. We often speak about the normal form. Before we solve our uniqueness problem, let us see the general idea on constructing normal forms Definition. Let {g i i J, J an arbitrary index set} = G K X and G =: I. Let τ { i : X K X : A(lm(g i ) lc(g i ) 1 g i )B, if x = Alm(g i )B for some A, B X x x otherwise and let τ i : K X K X be the K-linear continuation of τ. One calls τ i a reduction with g i. 8

9 Let f K X. One says that τ i acts trivially on f, if the coefficient of Alm(g i )B is zero in f for all A, B X. f is called irreducible, if all reductions act trivially on f. In other words τ i (f) = f i J Algorithm. Input: An ideal I K X with a given generating set G = {g i i J}, f K X Output: g, a reduced normal form of f w.r.t. G Set g = f. while τ i acts non-trivially on g for some i J do g = τ i (g); end while; return g; 1.12 Remark. By Definition 1.9, it is still not clear that the normal form is unique and in fact it is not. This is due to the fact that G is an arbitrary generating set and the construction of the normal form given in 1.11 depends on the choice of the reductor. Furthermore it is not guaranteed that Algorithm 1.11 will terminate. We have to make further assumptions Definition. A finite sequence of reductions τ i1,...,τ im is said to be final on f K X, if τ im τ i1 (f) is irreducible. An element f K X is called reduction-finite, if for any sequence {t ij } j=1 of reductions there exists m N, such that τ ik acts trivially on τ im τ i1 (f) for every k > m. An element f K X is called reduction-unique, if it is reduction-finite and if its images under all final sequences are the same Example. It is a good idea to see what happens in the commutative case, since it is a natural special case. So assume R = K[x 1,...,x n ]. All the definitions from above can be imported to R (in fact this is true for any sub- or factor algebra of K X, see for example [GP02]). Due to Hilbert s basis theorem R is Noetherian, that is, every ascending chain of ideals becomes stationary. Since for every f R we have deg(τ(f)) deg(f) for any reduction τ. So for any given sequence of reductions {τ i } i=1 with respect to G we have an ascending chain of ideals f + G τ 1 (f) + G τ 2 τ 1 (f) + G..., which 9

10 becomes stationary, that is there exists m N, such that τ i m τ i1 (f)+g = τ im τ i1 (f) + G m m. So any f R is reduction-finite. Furthermore f is reduction-unique, if and only if G is a Gröbner basis. This is due to the fact that every remainder after division with G equals zero Lemma. Let f K X be reduction-finite. Then Algorithm 1.11 returns a reduced normal form f of f after finitely many steps. If f is even reduction-unique, then its normal form does not depend on the choices we have to make during the computation. Proof: The termination of the algorithm is obvious. We have to show that f is in fact a normal form of f. Therefore we have to show that f τ(f) I for any reduction τ, because then f f I = G. Because of the definition of τ it is sufficient to prove the statement for monomials, so assume f is a monomial. If τ(f) = f there is nothing to prove, so assume otherwise, that is f = Alm(g)B for some A, B X, g G. Since K is a field we may assume g is monic, as stated before. Therefore we have: f τ(f) = Alm(g)B τ(alm(g)b) = Alm(g)B A(lm(g) g)b = A(lm(g) lm(g) + g)b = AgB g I. Assume f is not a reduced normal form, that is there exists g G, such that lm(g) t for some monomial t occurring non-trivially in f. But then t = Alm(g)B for some A, B X and the reduction τ g acts non-trivially on f, which is a contradiction. The last statement is clear by definition of reduction-uniqueness. Now we have the potential to compute normal forms, but it is somehow vague, since we have to make many choices and cannot be sure about the uniqueness. Therefore we want to find a special Gröbner basis, such that the choices we have to make are minimal. First we note that a Gröbner basis is a special generating set Lemma. Let G be a Gröbner basis of a given ideal I. Then I = G. Proof: Since G I we have G I, so take f I \ G with minimal degree, that is f := min { f I \ G } (the minimum exits because we assume deg( f) that < is a monomial ordering) and say without loss of generality that f is monic. By the definition of a Gröbner basis there exists g G such that lm(g) lm(f), say lm(f) = Alm(g)B for some A, B X. Then f = f AgB I and deg( f) < deg(f), so by minimality f G. But then f = AgB + f = AgB + a p pb p G, which is a contradiction. p P G 10

11 1.17 Definition. Let G K X and G =: I. G is called simplified or minimal, if lm(g) / L(G \ {g}) g G. G is called reduced Gröbner basis, if G is simplified, a Gröbner basis and for every g G we have: 1. g is monic. 2. g lm(g) is in reduced normal form with respect to I Remark. Note that we build the normal form with respect to I. This is only a technical issue: in fact it would be absolutely equivalent, if we had demanded a normal form with respect to G, since a Gröbner basis is a generating set and if a monomial is divisible by some leading monomial of a polynomial contained in I, then it is divisible by a leading monomial of an element of the Gröbner basis. However, with this formulation the reduction of g lm(g) does not depend on the choice of the Gröbner basis Theorem. Fix an ordering. For any ideal I K X consisting only of reduction-finite elements there exists a unique reduced Gröbner basis. Proof: Existence: Let F be an arbitrary Gröbner basis. Without loss of generality we assume that all elements of F are monic. If F is not simplified, there exists f F, such that lm(f) L(F \ {f}), that is, F \ {f} is still a Gröbner basis. By iterating this step we find a simplified Gröbner basis after a countable number of steps. Assume now F is a monic, simplified Gröbner basis and take f F. If f lm(f) is in reduced normal form we are finished. Otherwise we use Algorithm 1.11 to get an element f, which is the reduced normal form of f lm(f). (Note that the algorithm terminates, because I consists only of reduction-finite elements). Replace f by g := lm(f) + f and call the new set G. Then G is a simplified Gröbner basis, since lm(g) = lm(f). If we do this iteratively we get a reduced Gröbner basis G after a countable number of steps. Uniqueness: Let G, G be two reduced Gröbner bases. Take g G \ G. Since G is a Gröbner basis of I we have lm( g) L( G ) and g = a f fb f for some f f G. Assume lm( g) / {lm(g) g G}. Then there exists g G, such that lm(g) lm( g). But because G is a Gröbner basis as well there exists g G, such that lm( g) lm(f), which implies that lm( g) lm( g), and therefore, since G is reduced, we have lm( g) = lm(f) = lm( g). Repeating 11

12 this step for an g G \ G we get {lm( g) g G} = {lm(g) g G}. Take g G, g G, such that lm(g) = lm( g). Because g g I there exists f G, such that lm(f) lm( g g). Because of deg( g g) < deg(g) we have f g and there exists g f G, such that lm( f) = lm(f). Since lm(f) does not divide any term of g, lm( g g) must be a term occurring in g, say t. But then lm( f) = lm(f) t, a contradiction to the assumption that G is reduced Corollary. Let G be a simplified Gröbner basis of I K X consisting only of reduction-finite elements. Then all elements of {g lm(g) g G} are already reduction-unique. Proof: This is a immediate consequence of Theorem Remark. The question arises, whether all elements of K X are reductionunique with respect to a given reduced Gröbner basis, which would imply the existence of a unique normal form. The answer is yes, but to prove this we need some further knowledge. 1.3 The Gröbner Basis Algorithm For this section we will always assume that our ideal I is finitely generated, due to the fact that we want to do some computations, which would be quite difficult if we start with an infinite generating set. Nevertheless this assumption is not necessary. Note that even with a finite generating set we may get a Gröbner basis which is infinite (see 1.38). Again we may assume that all polynomials in a generating set are monic Definition. Let G = {g 1,...,g ω } K X. We call a polynomial f weak with respect to G, if f = ω k=1 c k,j l k,j g k r k,j, where c k,j K and l k,j, r k,j X such that l k,j lm(g k )r k,j lm(f) k = 1,...,ω. Let H K X. A polynomial f is called reducible from H with respect to G, if weakness with respect to G of all elements of H implies weakness of f with respect to G. Note that weakness is a special form of generating f with elements of G. Since it is allowed to use the same generator more than one time it should be allowed for weakness as well. For example the polynomial p := xy + yx + xyx y should be weak with respect to {y}. If one wants to avoid the twin-sum in the definition of weakness one can consider the enveloping algebra K X K X op, where K X op denotes the opposite algebra, that is, K X endowed with the multiplication a b = b a a, b K X. j 12

13 Then K X is a K X K X op module and the action of K X K X op on K X is given by: K X K X op K X K X : (l r, p) l p r 1.23 Definition. Let G = {g i 1 i ω} be a set of monic polynomials. An obstruction of G is a six-tuple (l, i, r; λ, j, ρ) with 1 i, j ω and l, r, λ, ρ X such that lm(g i ) lm(g j ) and llm(g i )r = λlm(g j )ρ. For any given obstruction we define the corresponding S-polynomial as s(l, i, r; λ, j, ρ) = lg i r λg j ρ. A set D of polynomials is called basic for G if every S-polynomial of G is reducible from D with respect to G Motivation. Starting with a generating set for I the set of all non-weak S-polynomials will be a Gröbner basis. This seems to be an easy way to compute a Gröbner basis, since one only has to compute all S-polynomials and check if they are weak or not. This procedure has the disadvantage that it would take forever, literally, since the set of all obstructions is infinite. So our medium-term issue is to discard most of these obstructions Lemma. Let G = {g i 1 i ω} be a set of monic polynomials and (l, i, r; λ, j, ρ) a weak obstruction, that is, the corresponding S-polynomial is weak with respect to G. Then all obstructions ( l, i, r; λ, j, ρ) with l = w 1 l, r = rw 2, λ = w 1 λ and ρ = ρw 2, where w 1, w 2 are arbitrary monomials, are also weak. Proof: Set s := s(l, i, r; λ, j, ρ) and s := s( l, i, r; λ, j, ρ). Because the obstruction is weak we can write s = lg i r λg j ρ = ω c k,l l k,l g k r k,l with c k,l K,l k,l, r k,l X, k=1 l k,l lm(g k )r k,l lm(s) k = 1,...,ω. Now we have s = lg i r λg j ρ = w 1 (lg i r λg j ρ)w 2 = w 1 sw 2 = w 1 ( ω c k,l l k,l g k r k,l )w 2 = ω c k,l lk,l g k r k,l with k=1 l lk,l = w 1 l k,l and r k,l = r k,l w 2. Furthermore we see that l k,l lm(g k ) r k,l w 1 lm(s)w 2 = lm( s), which shows that s is weak with respect to G. So multiples of obstructions need not be considered. However the set we have to consider is still infinite. But the lemma helps us to prove our claim in Theorem. For a set G of polynomials generating an ideal I of K X, the following statements are equivalent: (i) G is a Gröbner basis. (ii) The reduced normal form of each polynomial in I is equal to 0. l k=1 (iii) Each S-polynomial of G is weak with respect to G. l 13

14 (iv) The empty set is a basic set for G. Proof: (i) = (ii): Induction with respect to the monomial ordering <: The normal form of 0 equals 0. Take 0 f I and assume f is monic. Since G is a Gröbner basis there exists g G such that lm(g) lm(f), that is, l, r X : llm(g)r = lm(f). Because of f, g I we have f := f lgr I and deg( f) < deg(f). By the induction hypothesis, the normal form of f equals zero and we obtain that the normal form of f equals zero as well. (ii) = (iii): Suppose s = s(l, i, r; λ, j, ρ). By assumption the normal form of s with respect to G equals 0, so s is weak by the definition of weakness. (iii) (iv): Clear by definition. (iii) = (i): Suppose f I, but lm(f) / {lm(g) g G} and lm(f) is minimal with respect to <. Now there are at least two polynomials g i, g j G, g i g j, such that f = c i,l l i,l g i r i,l + c j,l l j,l g j r j,l + c k,l l k,l g k r k,l, l l g k G,g k g i,g j l c k,l K, l k,l, r k,l X k and t := lm( l i,l g i r i,l ) = lm( l j,l g j r j,l ) > lm(f). l l Now by assumption s := s(lm(l i,l ), i, lm(r i,l ); lm(l j,l ), j, lm(r j,l )) is weak and s = a k,l g k b k,l, where J is an arbitrary set of indices and g k G, such that all k J l leading terms of g k are smaller than t. Then f = lc(l i,l r i,l )lc(l j,l r j,l ) 1 l j,l g j r j,l + l lc(l i,l r i,l ) a k,l g k b k,l + l h,l g h r h,l is an expression of f with fewer summands with leading term equal to t. If we do this iteratively until we have only l k J h i,j l one term equal to t left, we reach a contradiction and we can conclude that G is a Gröbner basis. Note that the generating set is not taken to be finite. If we do not enumerate the polynomials in a generating set G, we often write (l, g, r; λ, p, ρ) for the obstruction of g, p G. Now we focus on finding a finite set of obstructions, from which we can construct a Gröbner basis. Therefore we introduce the concept of overlap Definition. We say two monomials t 1, t 2 X have overlap b X or overlap at b X if there are a, c X such that t 1 = ab and t 2 = bc or t 1 = ba and t 2 = cb or t 1 = b and t 2 = abc. If 1 is the only overlap between t 1 and t 2 we say the monomials have no overlap. Equivalently the monomials are called coprime An obstruction (l, i, r; λ, j, ρ) is said to have no overlap if there exists w X such that llm(g i )r = llm(g i )wlm(g j )ρ or llm(g i )r = λlm(g j )wlm(g i )r Example. Suppose g 1, g 2 K x 1, x 2, x 3 with lm(g 1 ) = x 1 x 2 and lm(g 2 ) = x 2 x 3. Then the only overlap between these monomials is x 2. It is easy to see, that the obstruction (1, 1, x α 2 x 3; x 1 x α 2, 2, 1) has no overlap (take w = xα 1 2 ). The more important question arises: Is the converse true? 14

15 1.29 Lemma (Product Criterion). Let g 1, g 2 K X be such that l 1 := lm(g 1 ) and l 2 := lm(g 2 ) have no overlap. Then every obstruction (l, g 1, r; λ, g 2, ρ) with l, r, λ, ρ X has no overlap. Proof: Since l 1 and l 2 have no overlap lm(lg 1 r) = lm(λg 2 ρ) implies that either ll 1 and λ or l 1 r and ρ have overlap l 1. Assume the first case is true. Then r and l 2 overlap at l 2, say r = l 2 r. Then r = ρ and therefore ll 1 r = ll 1 l 2 r = ll 1 l 2 ρ which shows that (l, g 1, r; λ, g 2, ρ) has no overlap. Now if l 1 r and ρ overlap at l 1 then l and λl 2 have overlap l 2 and l = ll 2 = λl 2. Hence we get ll 1 r = λl 2 l 1 r and again we obtain that (l, g 1, r; λ, g 2, ρ) has no overlap Theorem. Let G = {g i i = 1,..., ω} K X. Every obstruction without overlap is reducible from an S-polynomial with overlap with respect to G. Proof: Let b = (l, i, r; λ, j, ρ) be an obstruction without overlap and denote by s its S-polynomial. Since llm(g i )r = λlm(g j )ρ we have either r = wlm(g j )ρ or l = λlm(g i )w. If the former is valid then we also have λ = llm(g i )w and by Lemma 1.25 b = (l, i, wlm(g j )ρ; llm(g i )w, j, ρ) is reducible from (1, i, wlm(g j ); lm(g i )w, j, 1). Therefore we assume l = ρ = 1. Write g i = c h t h, g j = d p u p with t h, u p X, c h, d p K \ {0}, such that h p t h > t h+1 and u p > u p+1. Now s = g i r λg j = g i wlm(g j ) lm(g i )wg j = g i w(g j d p u p ) (g i c h t h )wg j = c h t h wg j d p g i wu p. Assume p,p 1 h,h 1 h,h 1 p,p 1 c 2 t 2 wu 1 = d 2 t 1 wu 2, that is the leading terms t 2 wlm(g j ) and lm(g i )wu 2 of the two summations cancel each other. Since t 2 < t 1 and u 2 < u 1 this only occurs if c 2 = d 2 and there are v 1, v 2 X, such that t 1 = t 2 v 1 and u 1 = v 2 u 2 with v 1 w = wv 2. If w is a left divisor of v 1, say v 1 = wv 1, then v 2 = v 2 w, which implies that v 1 = v 2 and therefore (1, i, wlm(g j ); lm(g i )w, j, 1) is reducible from (1, i, v 1 lm(g j); lm(g i )v 1, j, 1) by Lemma If w is not a left divisor of v 1, then w has a selfoverlap, that is, w = v 1 w = w v 2. and again we apply Lemma So we may assume w = 1 that is, b = (1, i, lm(g j ); lm(g i ), j, 1). We find s = g i lm(g j ) lm(g i )g j = lm(g i )lm(g j ) + p,p 1 h,h 1 p,p 1 h,h 1 c h t h lm(g j ) lm(g i )lm(g j ) lm(g i )d p u p = c h t h (g j d p u p ) (g i c h t h )d p u p = ( h,h 1 p,p 1 c h t h )g j g i ( d p u p ) g i, g j, p,p 1 h,h 1 so s is weak with respect to G, which implies that it is reducible from G. 15

16 The theorem states: If an S-polynomial s(l, g i, r; λ, g j, ρ) is not weak with respect to G, then the leading monomials of the two polynomials g i and g j have an overlap. This will help us to find a finite basic set Lemma. Let G = {g i i = 1,...,ω} K X. There is a finite basic set D of S-polynomials of G, such that every S-polynomial of G in D corresponds to an obstruction (l, i, r; λ, j, ρ) with overlap and with either one of the two parameters {l, λ} and one of {r, ρ} equal to 1 or λ = ρ = 1. Proof: We write s = s(l, i, r; λ, j, ρ), lm(g i ) = m 1...m p and lm(g j ) = n 1...n q with m k, n k X of degree 1, k = 1,...,p; k = 1,..., q (this means that each m k and n k corresponds to an x i, i = 1,..., n). Now if s is not weak, then it must have some overlap. In particular, lm(g i ) and lm(g j ) must overlap. This can occur in three ways: m 1 m h = n q h+1 n q, 1 h < p, n 1 n h = m p h+1 m p, 1 h < p, m 1 m p = n h+1 n h+p, 1 h < q p. In particular, for every two polynomials the number of possible overlaps is finite. We show that D needs to contain at most one S-polynomial for every overlap, which completes the proof. Assume lm(g i ) and lm(g j ) have nontrivial overlap. To satisfy the equation llm(g i )r = λlm(g j )ρ, the factors that are not in the overlap have to be in λ or ρ respectively in l or r (cf. proof of Lemma 1.29). So for every obstruction corresponding to some overlap the monomials llm(g i )r and λlm(g j )ρ have to be equal to lw r and λw ρ, respectively, with w equal to w = n 1 n q h lm(g i ) = lm(g j )m h+1 m p, w = lm(g i )n h+1 n q = m 1 m p h lm(g j ), w = n 1 n h lm(g i )n h+p+1 n q = lm(g j ), in the respective cases. Now by Lemma 1.25 these obstructions are weak except when l = r = λ = ρ = 1. So for every possible overlap there exists a single S-polynomial such that all other obstructions are reducible from it with respect to {g i, g j }; in the respective cases, the corresponding obstructions are (n 1 n q h, i, 1; 1, j, m h+1 m p ), (1, i, n h+1 n q ; m 1 m p h, j, 1), (n 1 n h, i, n h+p+1 n q ; 1, j, 1). This means that s need only to be in D if at least one of the two parameters l and λ and one of the two parameters r and ρ are equal to 1. We refer to the S-polynomials of g and g, we have to consider, as S(g, g). We distinguish between three kinds of obstructions: 16

17 1.32 Definition. Let s = (l, i, r; λ, j, ρ) be an obstruction of the set G = {g i 1 i ω} of monic polynomials in K X. If l = 1, then we call s a right obstruction. If l 1 and r = 1, then we call s a left obstruction. If s is not a right nor a left obstruction and λ = ρ = 1, then we call s a central obstruction Corollary. Let G be a set of polynomials in K X and let D be the set of all non-zero normal forms of S-polynomials with respect to G corresponding to all left, right and central obstructions of G. Then D is a basic set for G. Proof: This is exactly the statement of In Definition 1.32 the restriction to a finite set G is not necessary, since an obstruction includes only two polynomials. However, as stated before, for reallife computations finiteness is required and so we will assume for the rest of this section that G = {g i 1 i ω}. The overlaps given in 1.32 are also called ambiguities, since Bergman used this term in his famous work [Ber78]. Because one of the goals of this work is to translate the Diamond Lemma into a modern language we will stick to the term overlap. But before we come to this matter we introduce an algorithm that computes a reduced Gröbner basis Definition. Let I be a two-sided ideal of K X and let G, D be subsets of K X. We say that (G,D) is a partial Gröbner pair for I if the following properties are satisfied: 1. All polynomials in G D are monic. 2. G is a generating set of I. 3. Every element of D belongs to I and it is in normal form with respect to the polynomials in G. 4. The set D is basic for G. 5. For every f G the normal form with respect to G D of the normal form with respect to G \ {f} equals zero Remark. Let I be a two-sided ideal in K X and let (G,D) be a partial Gröbner pair for I. If D is the empty set, then G is a Gröbner basis. Since K X is not Noetherian, for example the ideal x 1 x n 2x 1 n N can not be finitely generated, our algorithm may not terminate in all cases. However, we will see later that we can use this algorithm to get some important results after finitely many steps. 17

18 1.36 Algorithm. Input: a (finite) generating set G for I K X Output: a reduced Gröbner basis for I Compute all non-zero normal forms of S-polynomials with respect to G corresponding to all left, right and central obstructions of G and call the resulting set D. Then (G,D) is a partial Gröbner pair. Construct a new partial Gröbner pair ( G, D) as follows: 1. Take f D and set G = {g 1,...,g ω, g ω+1 := f}. 2. Compute the left, right and central obstructions of G of the form (l, i, r; λ, ω + 1, ρ) and (l, ω + 1, r; λ, j, ρ) for certain i, j {1,...,ω} and l, r, λ, ρ X and put the non-zero normal forms of their S-polynomials with respect to G D in D, such that D becomes a basic set for G. Call this new basic set D. 3. For each i {1,..., ω} compute the normal form g i with respect to G \ {g i } of g i. If g i = 0 remove g i from G. Otherwise, if g i is distinct from g i, a) replace g i by g i ; b) compute the left, right and central obstructions of the new G involving g i; c) if the normal form with respect to G D of an S-polynomial of such an obstruction is non-zero then add its normal form to D. 4. Replace each d D by its normal form with respect to ( G D) \ {d} Theorem. In the situation of 1.36, the ideal generated by the leading monomials of G is strictly contained in the ideal generated by the leading monomials of G. If D = then G is a Gröbner basis for I (and the routine stops). Proof: First we have to show that ( G, D) is a partial Gröbner pair, which means we have to verify condition one to five of Definition Since all polynomials in G and D are normal forms, they are monic, we get condition 1. If g i G adjusted as in step 4 of the algorithm, then the ideal generated by {g i} (G \ {g i }) coincides with I, so we get condition 2. Clearly all elements of D belong to I and are in normal form with respect to G and this is condition 3. Because of 1.33, D is a basic set for G and hence condition 4. For every element g G \ G, the normal forms of the newly computed central obstructions of G involving g take care of condition 5. That L(G) L( G) is valid follows immediately from the construction we have made. The final assertion is a consequence of Remark

19 1.38 Example. For all examples we take the lexicographical ordering with x 1 > x 2 >... > x n or x > y > z respectively. Take K x, y and G 1 = {xyx + y 2 }. There is only one obstruction to consider, since the only central obstruction are the trivial ones and every left obstruction is equal to a right obstruction, namely (xy, 1, 1; 1, 1, yx) = xy 3 y 3 x. = D 1 = {xy 3 y 3 x}. Now G 2 = {xyx + y 2, xy 3 y 3 x}, since xy 3 y 3 x is in normal form with respect to g 1. Because our new g 2 only has trivial obstruction with itself, there is only one new obstruction: (1, 1, y 3 ; xy, 2, 1) = y 5 + xy 4 x, which has normal form 0 with respect to G 2, so G 2 is a Gröbner basis for I = G 1. Take G = {x i x j x j x i 1 i < j n} K X. We claim that G is already a Gröbner basis. The only non-trivial overlaps are given by the polynomials x i x j x j x i and x j x w x w x j, where 1 i < j < w n. The S-polynomial can be computed by (x i x j x j x i )x w x i (x j x w x w x j ) = x i x w x j x j x i x w which reduces to zero, using the leading monomials of x i x w x w x i, x j x w x w x j and x i x j x j x i G. Note that G generates the commutator ideal, so we have K[x 1,...,n] = K X / G. Let us consider the generating set B = {yzxy xyzx, zxyz xyzx, zxyz yzxy} K x, y, z, which consists of braid relations (cf. [Gar07]). Then the unique reduced Gröbner basis is given by G = {yzxy zxyz, xyzx zxyz, xzxyz zxyzy, yz n xyz zxyz 2 x n 1, xz n xyz zxyzyx n 1 n N}. Obviously, none of the elements of G is redundant. To see that G is in fact a Gröbner basis one has to consider all pairs (g i, g j ) of elements of G and check if all possible obstructions of (g i, g j ) vanish to zero. We demonstrate this for w 1 := yz n xyz zxyz 2 x n 1 and w 2 := yz m xyz zxyz 2 x m 1 for arbitrary n, m N. We only have to worry about the right overlap, since n and m are arbitrary elements in N (so we can exchange their places for the left overlap). Now w 1 and w 2 overlap at yz and we have: 19

20 (yz n xyz zxyz 2 x n 1 ) z m 1 xyz yz n x (yz m xyz zxyz 2 x m 1 ) = zxyz 2 x n 1 z m 1 xyz + yz n xzxyz 2 x m 1 xzxyz zxyzy yz n+1 xyzyzx m 1 zxyz 2 x n 1 z m 1 xyz yz n+1 xyz zxyz 2 x n xyzx zxyz xyzx zxyz xz 2 xyz zxyzyx xzxyz zxyzy zxyz 2 x n yzx m 1 zxyz 2 x n 1 z m 1 xyz zxyz 2 x n 1 zxyzx m 2 zxyz 2 x n 1 z m 1 xyz zxyz 2 x n 1 z 2 xyzx m 3 zxyz 2 x n 1 z m 1 xyz zxyz 2 x n 2 zxyzyx m 2 zxyz 2 x n 1 z m 1 xyz zxyz 2 x n 3 zxyzy 2 x m 2 zxyz 2 x n 1 z m 1 xyz xz m 1 xyz zxyzyx m 2 xzxyz zxyzy 0. zxyz 2 x n 3 zxyzy 2 x m 2 zxyz 2 x n 2 zxyzyx m The Diamond Lemma We now state our version of the Diamond Lemma, which will give us a uniquely determined normal form in certain situations. We will see that the assumptions we have to make are mostly for ensuring the existence of a reduced Gröbner basis Definition. An ordering < is said to fulfill the descending chain condition if every descending chain of monomials (with respect to <) becomes stationary. Equivalently one says that < is Artinian or well-founded. Note that if < fulfills the descending chain condition every element of K X is reduction-finite. Recall that we always assume we have a monomial ordering, in particular, we have a well-ordering, which implies that the ordering is Artinian Lemma. For a given subset G K X we have: (i) The set of reduction-unique elements of K X (w.r.t. G) forms a K-subspace of K X and we have an K-linear map r G from this subspace into K X irr, the set of all irreducible elements of K X. (ii) Suppose a, b, c K X are such that for all monomials A, B, C occurring with non-zero coefficient in a, b, c, respectively, the product ABC is reduction-unique. (In particular this implies that abc is reduction-unique.) Let r be any finite composition of reductions. Then ar(b)c is reductionunique, and r G (ar(b)c) = r G (abc). 20

21 Proof: (i) Define r G as the linear continuation of the map, that maps any given reduction-unique element to its uniquely determined reduced normal form. Let a, b K X be reduction-unique and take k K. Then ka + b is reduction-finite, since reductions are linear maps and a and b are reductionfinite. Let r be a composition of reductions, such that r is finite on ka + b. Since a is reduction-unique there exists a composition with reduction r, such that r r(a) = r G (a) and similar there is r, such that r r r(b) = r G (b). Because r(ka + b) is irreducible, we have r(ka + b) = r r r(ka + b) = k r r r(a) + r r r(b) = kr G (a) + r G (b) and our claim follows. (ii) By (i) it suffices to prove the claim for a, b, c X and a single reduction r. But then we have ar(b)c = r(abc) and hence it is reduction-unique if and only if abc is, with the same reduced normal form. Again the only challenge we meet is given by monomials involving an overlap: Assume we have three monomials A, B, C and consider AB and BC, such that these monomials have overlap B. Recall that reductions were defined for monomials. Therefore there might exist a reduction for AB and a different one for BC (assume A 1 C), say τ and σ. Then τ(abc) σ(abc) and we have two different ways to reduce ABC Definition. Assume A, B, C X. Consider AB and BC and let τ be a reduction on AB and σ be an reduction on BC. The overlap is called resolvable if there exist two compositions of reductions r and r, such that r(τ(abc)) = r (σ(abc)). This is also known as the diamond condition Remark. The name diamond condition is taken from the field of graph theory, where the Diamond Lemma was stated first. It refers to the fact that for every two different edges τ, σ starting from the same vertex v there will be paths r and r such that r τ(v) = r σ(v), thus the paths forming a diamond as illustrated in Figure 1.1. But since graphs will be our matter in the next chapter we stick to the formulation within terms of the free algebra Definition. Assume G K X. The set of all reductions defined by G (cf. 1.10) is called reduction system. We refer to the overlaps occurring in the leading monomials of G as the overlap of the reduction system G. This is merely a renaming. The intention is rather obvious: With a given Gröbner basis G we want to reduce all polynomials to a normal form, therefore we may call G reduction system. The Diamond Lemma gives us now a condition for 21

22 ABC τ σ p 1 p 2 r r nf Figure 1.1: The Diamond Graph uniqueness of the reduced normal form, namely the diamond condition. Recall that we have defined the normal form with respect to an arbitrary subset of K X. However in general the term reduction system depends on the chosen ordering: If we have two different orderings < and on K X the leading monomials of G with respect to < may be different from the ones with respect to and hence we have different reductions. Luckily, the Diamond Lemma states that if the overlaps are resolvable with respect to one Artinian ordering they are resolvable with respect to any Artinian ordering Theorem (Diamond Lemma). Let G be a reduction system and < an Artinian ordering. Then the following conditions are equivalent: (i) All overlaps of G are resolvable. (ii) All overlaps of G are resolvable with respect to <. (iii) All elements of K X are reduction-unique under G. Proof: Since the implications (iii) (i) (ii) are obvious we only have to prove (ii) (iii). So assume (ii). Because the reduction-unique elements form an ideal of K X we only have to prove our claim for monomials D X. This is done by induction over the degree of D (with respect to <). For deg(d) = 0 there is nothing to show, so assume all monomials with degree less than that of D are reduction-unique, that is K X <D := {f K X deg(f) < deg(d)} Im(r G ). Let r, r be two reductions acting non-trivially on D. We want to show r G (r(d)) = r G (r (D)). Assume D = LABCM, r = r AB and r = r BC, which corresponds to the case that the monomials we use to reduce D have a right overlap. Then 22

23 we have r(d) r (D) = L(f AB C Af BC )M, where f m is the image of the monomial m under the corresponding reduction. By condition (ii) we have f AB C Af BC I ABC, where I E, E X, denotes the ideal of K X spanned by all elements F ph, F, H X, p K X, such that F lm(p)h < E. Therefore we have L(f AB C Af BC )M I D. By assumption I D is annihilated by r G, so we have r G (r(d) r (D)) = r G (r(d)) r G (r (D)) = 0 as required. The case D = LABCM, r = r B and r = r ABC is completely analogous. Finally, let D = LABCM, r = r A and r = r C, where A C are disjoint words, that is, the monomials have no overlap at all. By Lemma 1.40 (ii) we know that r G (Lf A BCM) = r G (LABf C M), which completes the proof Corollary. Let G be a reduced Gröbner basis and < an Artinian ordering. Then for every element of K X there exists a unique reduced normal form with respect to G. Proof: Let g 1, g 2 G such that m 1 := lm(g 1 ) = AB, m 2 := lm(g 2 ) = BC for some monomials A < B < C X, that is, g 1 and g 2 have an overlap (due to the fact that G is reduced this is the only overlap that can occur). Then we have Ag 2 g 1 C := g 3 G or g 3 = 0. Denote by τ i the reduction with g i. Assume g 3 = 0. Then we have τ 1 (ABC) τ 2 (ABC) = m 1 C g 1 C Am 2 + Ag 2 = m 1 C Am }{{} 2 + Ag }{{} 2 g 1 C = 0. So the overlap is resolvable. }{{} =ABC =ABC =0 Assume g 3 G. Again we get τ 1 (ABC) τ 2 (ABC) = m 1 C g 1 C Am 2 + Ag 2 = Ag 2 g 1 C which can be reduced to zero by g }{{} 3, showing that this =g 3 overlap is resolvable. Now 1.44 is applicable and our claim is proven Remark. So we got the uniqueness of the reduced normal form. Note that with an Artinian ordering, we will always have a reduced normal form and may apply Algorithm 1.11 to compute it. However, even with this setup our reduced Gröbner basis may be infinite and Algorithm 1.11 has to check infinitely many reductions, even though only finitely many of them act non-trivially. So termination is not guaranteed, not to mention that after finitely many steps does not imply computable in an acceptable amount of time. Some tricks to deal with this are presented in the next section. 23

24 1.5 Truncated Gröbner Bases Our goal for this section is to see how much information we can gather out of a part of a Gröbner basis. For this we first have to define what a part of a Gröbner basis is Definition. Let G be a set of polynomials such that deg t (g) q g G and some q N. In Algorithm 1.36 discard every obstruction with an S- polynomial of total degree greater than q. If the algorithm returns the set G q, we call G q a truncated Gröbner basis of degree q. Let B be a Gröbner basis for G and G G. We call G a partial Gröbner basis, if it is already a Gröbner basis for the ideal Ĩ := G. Since 1.36 always computes a reduced Gröbner basis, a truncated Gröbner basis will always be reduced. Note that a truncated Gröbner basis does not necessarily need to be a subset of our reduced Gröbner basis. But since the algebra K X has only finitely many variables there are only finitely many monomials of total degree q (up to scaling), so the truncated version of the algorithm will terminate. It is clear, that G q = G, since Algorithm 1.36 does not change the generated ideal. So we may use G q to get to know more about the Gröbner basis we want to compute Lemma. Let B K X and G q be a truncated Gröbner basis of degree q of B. If max{deg t (g) g G q } q 2 then G q is a Gröbner basis of the ideal generated by B. Proof: Define m := max{deg t (g) g G q }. Since B = G q we only need to show: Every S-polynomial of obstructions of polynomials in G q is of degree at most 2m 1, which implies the claim. Take g m G q such that deg t (g m ) = m. Take an arbitrary g i G q, such that (l, i, r; λ, m, ρ) is a left, right or central obstruction. Note that deg t (g i ) deg t (g m ) g i B, so all obstructions we need to consider are of the form (l, i, r; λ, m, ρ). Because of 1.30 we may assume that lm(g i ) and lm(g m ) have overlap b Assume lm(g i ) = ab and lm(g m ) = bc with deg t (b) 1. Clearly (1, i, c; a, m, 1) is an obstruction and the induced S-polynomial is of degree at most 2m 1, since b is not a constant. Let (1, i, r; λ, m, ρ) be a right obstruction. Since lm(g i r) = lm(λg m ρ) ablm(r) = lm(λ)bclm(ρ) we get lm(λ) = a and lm(r) and clm(ρ) have overlap c. So s(1, i, r; λ, m, ρ) = g i c r ag m ρ for some r, ρ K X, which is weak with respect to G q {s(1, i, c; a, m, 1)} by the definition of weakness. Now let (l, i, 1; λ, m, ρ) be a left obstruction. As before we get s(l, i, 1; λ, m, ρ) = lg i c λag m, which is weak with respect to G q {s(1, i, c; a, m, 1)}. By assumption there will not be any central obstruction. 24

25 2. The case g i = ba and g m = cb is completely analogous to part Because the degree of g m is maximal, the last case we have to study is g m = ag i b. But this would imply that g m is weak with respect to G q \ {g m } which is a contradiction to the assumption that G q is a truncated Gröbner basis Corollary. If G q is a truncated Gröbner basis, then H := {p G q deg t (p) q 2 } is a partial Gröbner basis for G q. Proof: Clear by Provided there exists a finite Gröbner basis, this leads to a way to compute the whole Gröbner basis starting with a truncated one, by iteratively increasing the degree bound Algorithm. Input: A (finite) truncated Gröbner basis G q for I = G q Output: A reduced Gröbner basis for I ( ) p := max{deg t (g) g G q } Apply the truncated version of Algorithm 1.36 to G q with degree bound 2p 1 and call the result G 2p 1 if p = max{deg t (g) g G 2p 1 } then return G 2p 1 else: go to ( ) end if 1.51 Remark. It is obvious that 1.50 terminates, if there exists a finite Gröbner basis, and that it will return this Gröbner basis of I. The proof of Lemma 1.48 states that if we construct an S-polynomial we will lose at least one degree to the overlap, since it is not trivial. This illustration shows us that our lemma includes only the worst case. In fact most of the time we will not have to double our q for the truncated Gröbner basis, as the following lemma states: 1.52 Lemma. Let B K X and G q be a truncated Gröbner basis of degree q of B. Take g 1 G q of degree m, and g 2 G q of maximal degree, say o, such that g 2 has a non-trivial and non-central overlap with g 1. Define l := lcm(lm(g 1 ), lm(g 2 )), where lcm denotes the least common multiple, that is lcm(lm(g 1 ), lm(g 2 )) := max deg t (b) {b X lm(g 1 ) and lm(g 2 ) have overlap b} and set p := deg t (l). Then m + 1 p m + o 1. Proof: Assume lm(g 1 ) = ab and lm(g 2 ) = bc for some a, b, c X, which corresponds to a right obstruction. Since the overlap is non-trivial, none of the monomials a, b, c equal one, so they are all of positive degree. Therefore l = abc is of degree p = deg t (abc) = deg t (g 1 ) + deg t (c) m + 1 on the one hand and on 25